# Thread: Projection of line on plane

1. ## Projection of line on plane

Given $\displaystyle \pi: \begin{cases}l_1: \frac{x-2}{4}\ = \frac{y-1}{2}\ = \frac{z+5}{-4} &\\l_2: \frac{x+4}{-2}\ = \frac{y+1}{0}\ = \frac{z}{1} & \end{cases}$ and the point $\displaystyle M=(1,2,3)$ outside the plane. Find the projection of the line $\displaystyle M_1P$ on plane $\displaystyle \pi$ where $\displaystyle P$ is the intersection point of lines $\displaystyle l_1, l_2$.

All I can do is that I can find the equation of plane $\displaystyle \pi$ but don't have any idea what to do next.

I need your help.

Thank you.

2. Originally Posted by patzer
Given $\displaystyle \pi: \begin{cases}l_1: \frac{x-2}{4}\ = \frac{y-1}{2}\ = \frac{z+5}{-4} &\\l_2: \frac{x+4}{-2}\ = \frac{y+1}{0}\ = \frac{z}{1} & \end{cases}$ and the point $\displaystyle M=(1,2,3)$ outside the plane. Find the projection of the line $\displaystyle M_1P$ on plane $\displaystyle \pi$ where $\displaystyle P$ is the intersection point of lines $\displaystyle l_1, l_2$.
First a comment: No competent textbook or instructor would write a like this one did. It should be $\displaystyle l_2: \frac{x+4}{-2}\ = \frac{z}{1};~y=-1.$.

Now I will not do the work for you.
First show the line intersect in $\displaystyle P$.
The directions of the lines are $\displaystyle D_1:<4,2,-4>~\&~D_2:<-2,0,1>$.
The plane containing $\displaystyle P$ and with normal $\displaystyle N=D_1\times D_2$ is $\displaystyle \pi$.
The projection is the intersection of the plane containing $\displaystyle P$ with normal $\displaystyle \overrightarrow {PM} \times N$ with plane $\displaystyle \pi.$